Find the Area Between the Curves y=x^2+5x-4 , y=x+3

Problem

y=x2+5*x−4,y=x+3

Solution

1. Find the intersection points by setting the two equations equal to each other to determine the limits of integration.

x2+5*x−4=x+3

2. Rearrange the equation into standard quadratic form to solve for x

x2+4*x−7=0

3. Apply the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a) to find the roots.

x=(−4±√(,4−4*(1)*(−7)))/(2*(1))

x=(−4±√(,16+28))/2

x=(−4±√(,44))/2

x=−2±√(,11)

4. Identify the upper and lower functions on the interval [−2−√(,11),−2+√(,11)] Testing a point like x=−2 shows that y=x+3 is the upper curve.

ƒ(x)=x+3

g(x)=x2+5*x−4

5. Set up the integral for the area A=(∫_a^b)((ƒ(x)−g(x))*d(x))

A=(∫_−2−√(,11)^−2+√(,11))(((x+3)−(x2+5*x−4))*d(x))

A=(∫_−2−√(,11)^−2+√(,11))((−x2−4*x+7)*d(x))

6. Evaluate the definite integral using the power rule.

A=[−(x3)/3−2*x2+7*x](−2+√(,11))(−2−√(,11))

7. Simplify the result using the property that for a quadratic a*x2+b*x+c with roots (x_1) and (x_2) the area is |a|/6*((x_2)−(x_1))3

(x_2)−(x_1)=(−2+√(,11))−(−2−√(,11))=2√(,11)

A=1/6*(2√(,11))3

A=1/6*(8⋅11√(,11))

A=(44√(,11))/3

Final Answer

Area=(44√(,11))/3

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